One mole of an ideal gas in initial state A undergoes a cyclic process ABCA as shown. Its pressure at A is Po. then
(a) Internal energies at A and B are same
(b) Work done by the gas in the process AB is PoVo ln4
(c) pressure at C is Po/4
(d) Temperature at C is To/4
shouldn't the answer be (a),(b),(c),(d)??? fiitjee says only a and b
39Why exactly won't the line pass through origin??
It's ideal conditions. PV = nRT.
If we plot a graph between V and T(obviously the other quantities are constant), we get a straight line of the form y = mx where m = nR/P...
Doesn't that pass through the origin???
If you think P isn't constant, that can't be. Had that been the case, we'd get a many sloped curve, not a straight line.
(Discussed this with govind. Any corrections are welcome)
1IIT PROFESSORS SAID THAT THOSE WHO WILL MARK ALL 4 CHOICES WILL BE CAUGHT BCOZ THEY THINK THAT THOSE BCOZ OF THAT PARTIAL MARKS TECHNIQUE THEY WILL MARK ALL 4 WHICH IS ABSOLUTELY ABSURD THIS QUESTION HAD REALL ABCD AS THE SOLUTION
1Option No. (A) and (B) ,here , seems easy to go through .
Regarding the (C) and (D) options ,there is a controversy going on about whether the line BC passes through origin or not.
Now,
Remember Charles Law - " At constant pressure , the volume of a given mass of gas is directly proportional to the absolute temperature of the gas " .
Another way to state the same is like - " Pressure remaining constant , the volume of a given mass of gas increases or decreases by 1/273 part of it's volume at 0°C for a rise or fall in temperature of 1°C "
i.e. volume of gas at t°C , V = V0(1 + t/273) , where V0 is the volume of the gas at 0°C .
Considering this expresion let us find the volume of gas at -273 °C i.e 0 K ,
V= V0(1 - 273/273) = 0
This implies , w.r.t the question , that the line BC (V/T = constant) must pass through the origin . Again, it is well known that ideal gases have zero volume at 0 K.
So, why aren't the coaching institutes paying a real attention to this fact ?
Do they have the doubt that whether the temperature on the x-axis is in °Centigrade or Kelvin ?
Obviously , it's in Kelvin , Ideal gas laws are valid for Kelvin scale only !
Also what Pritish and Subhomoy said earlier is correct and the same.
Hence i think the answers should be (A)(B)C)(D).
1IIT would give anwers jab mamla thanda ho jayega
jaisa sab bhool jaye tab tak
49ideal gases have 0 volume at 0 absolute temperature!!
so no y intercept!! since (0,0) is a point on graph!!
so
answer should be no way be a and b
correct set of answers must be a b c d
1very easy one
typical iit kind
A, B
is the asnwer
in the other procees neither slope is given nor any relation nor any cordinate
so unknown process are happneing
1WTF!!! did they mention in the optics question whether the line was passing out straight of the lens or not???wht did u do there????start writing coordinates.....dint u use ur eye estimation or a ruler to chek dat!!!
@ rajat : no offence....but if u can answer my query...how did u find the refractive index relations in the second paper match the column!!!
21i also marked ABCD in the exam but now i think it should be AB only.
1hey i also SAY THAT ANS IS A,B ,,,,,,, SINCE CD LINE MAY OR MAY NOT BE PASSING THROUGH ORIGIN....!!
33These are the questions IIT keep in the paper to make the ranking or marks unpredictable.. in these they can do anything...
There has been a case against the IITJEE selection procedure to make it more transparent. Lets see...
